On alternating products of graph relations

On alternating products of graph relations. Bodendiek Institut f. Mathematik und ihre Didaktik Christian-Albrechts-Universität Kiel.Lang Fachbereich Informatik Universität Hamburg Abstract It is well-known that one can give an elegant version of the Kuratowskitype theorem for the projective plane by means of the five elementary relations i, i = 0, 1,..., 4, on the set Γ of all finite, undirected graphs without loops and multiple edges. Furthermore, these five relations play an interesting role in didactics of mathematics. Following a theory given in [], C.Thies investigates them in []. In order to show that 0, 1,..., are an appropriate curriculum he has to deal with so-called alternating products T ( i ) T (i 1 ) T ( i ) T (i 1 )... or T (i 1 ) T ( i ) T (i 1 ) T ( i )..., i = 0, 1,..., 4, where T ( i ) denotes the reflexive transitive closure of i and T (i 1 ) = T ( i ) 1 the reflexive, transitive closure of i 1. Here, it is shown that, in case of i = 0, there exists exactly one alternating product in the set of all alternating products of 0, in case of i = and i = 4 the sets of all alternating products of and are infinite sets. 1

1 Introduction In the theory of embedding graphs into orientable and nonorientable surfaces, the five elementary relations 0, 1,...,, introduced in [1], play an important role because it is possible to give elegant and precise definitions of the subgraph relation and the subdivision relation and the minor relation on the set Γ of all finite, undirected graphs without loops and multiple edges. Observing that an ordered pair (, H) Γ Γ belongs to 0 iff H arises from by removing an edge e E() or an isolated vertex v V () the subgraph relation is equal to the reflexive, transitive closure T ( 0 ) of 0. An ordered pair (, H) Γ Γ belongs to 1 iff H arises from by contracting an edge e = {v, v } E(), provided at least one of the two endpoints v, v has got degree 1 or, and v, v do not have a common neighbour in. If U 1 = 0 1, then the well-known subdivision relation is equal to the reflexive, transitive closure T (U 1 ) of U 1 = 0 1. To give a precise definition of the famous minor relation on Γ we need 0, 1 and. An ordered pair (, H) Γ Γ belongs to iff H arises from by the contraction of an edge e = {v, v } E() provided v, v do not have a common neighbour in, and both v and v have got a degree. v v 1 v 1 v 1 v v u 1 u u v u u 1 v Q v u X v u O Figure 1: If U denotes the union U = 0 1, then the minor relation on Γ is equal to the reflexive, transitive closure T (U ) of U = 0 1.

v 1 v v u u u 1 v u v P (P) (P) u 4 (P) (P) = K 6 Figure : v 1 v v e D u u1 v1 u v u=v T u u 1 Figure :

It is almost superflous to mention that T ( 0 ), T (U 1 ), T (U ) are partial ordering relations on Γ while 0, 1, are not partial ordering relations on Γ. Now, it remains to give the definition of and. The great advantage of and and their reflexive, transitive closures T (U ), T (U 4 ), U i = 0 1... i, i =, 4, is that we need essentially less graphs in order to characterize the set of all graphs not embeddable into an orientable or nonorientable surface. Due to [1], an ordered pair (, H) Γ Γ belongs to iff H arises from by substituting a star S = {v} {v 1, v, v }, v, v 1, v, v V (), by the triangle K = (v 1, v, v ) provided {v 1, v, v } is an independent set in V. Figure 1 shows that the octahedron O arises from the cube Q by applying twice. In other words, we can say that the ordered pair (Q, O), or equivalently (Q, O) T ( ). Figure shows a representation of the Petersen graph P and four other graphs (P ), (P ), (P ), (P ) in the projective plane such that it is clear that the ordered pair (P, K 6 ), or equivalently (P, K 6 ) T ( ). 1=u =v 1 1 u 1 v 1 u=6 v=5 u=v 4=v =u K, v u K 5 Figure 4: Due to [1] an ordered pair (, H) Γ Γ belongs to iff H arises from by substituting a double star D by a double triangle T (Figure ) provided 4

the edges {v 1, v } and {u 1, u } do not belong to. Figure 4 shows the decisive advantage of because the two Kuratowski graphs K 5 and K, have got the property that the ordered pair (K,, K 5 ). Alternating products In order to be able to define the concept of an alternating product it makes sense to introduce the converse relations of i, i = 0, 1,,, 4. Since we are going to restrict ourselves to the converse relations 0 1, 1 and 4 1 we omit the definitions of 1 1 and 1. Observing the definitions of 0, and we obtain a) an ordered pair (, H) Γ Γ belongs to 1 0 iff H arises from by adding an edge e P (V ) E or a new isolated vertex v / V, b) an ordered pair (, H) Γ Γ belongs to 1 iff H arises from by substituting a triangle C = (v 1, v, v ) by the star v {v 1, v, v } such that v is a new vertex, not belonging to, c) an ordered pair (, H) Γ Γ belongs to 1 4 iff H arises from by substituting a double triangle T (1-amalgamation of two disjoint triangles) by a double trihedral D (double star) (Figure ). Observing T (i 1 ) = T ( i ) 1, i = 0,, 4, we are able to introduce alternating products A i (n) and B i (n), i = 0,, 4, of n N, n, factors as follows { T (i ) T ( A i (n) = i ) 1... T ( i ) T ( i ) 1 if n 0 mod, T ( i ) T ( i ) 1... T ( i ) 1 T ( i ) if n 1 mod, { T (i ) B i (n) = 1 T ( i )... T ( i ) 1 T ( i ) if n 0 mod, T ( i ) 1 T ( i )... T ( i ) T ( i ) 1 if n 1 mod. If we denote the sets of all alternating products of i, i = 0,, 4, by A i and B i, defined as follows A i = {A i (n) n N {1}}, B i = {B i (n) n N {1}}, (1) we are going to investigate the question whether the sets A i and B i, i = 0,, 4, are finite or infinite. 5

The finiteness of A 0 and B 0 Since first investigations lead to suppose that A 0 = {A 0 (), A 0 (), A 0 (4),...} () and B 0 = {B 0 (), B 0 (), B 0 (4),...} () are finite we first restrict ourselves to A 0 () and B 0 (), and prove Theorem 1 A 0 () = B 0 () = Γ Γ (4) T ( 0 ) T ( 0 ) 1 = T ( 0 ) 1 T ( 0 ) = Γ Γ. Proof: Because of T ( 0 ) T ( 0 ) 1 Γ Γ and T ( 0 ) 1 T ( 0 ) Γ Γ it suffices to show the two inclusions (a) and (b) by (a) T ( 0 ) T ( 0 ) 1 Γ Γ and (b) T ( 0 ) 1 T ( 0 ) Γ Γ. Ad (a) : Observing the definitions of a subset and a composition of relations the inclusion (a) is equivalent to the proposition [(, X) T ( 0 ) (X, H) T ( 0 ) 1 ]. (5) (,H) Γ Γ X Γ Assume (, H) is any ordered pair in Γ Γ. As the intersection graph H is a subgraph of both and H we obtain (, H) T ( 0 ) and (H, H) T ( 0 ) such that ( H, H) T ( 0 ) 1. Putting X = H we obtain (, H) T ( 0 ) and ( H, H) T ( 0 ) 1 such that (5) is true. This fact proves the inclusion (a). Ad (b): Similarly to (a) the inclusion (b) is equivalent to the proposition (6) by [(, Y ) T ( 0 ) 1 (Y, H) T ( 0 )]. (6) (,H) Γ Γ Y Γ 6

The proof of (6) is done by choosing Y = H(= union graph of and H) for any graphs, H Γ for and H are subgraphs of Y = H implying ( H, ) T ( 0 ), ( H, H) T ( 0 ) and (, H) T ( 0 ) 1. This proves (b). The answer of the question asked above concerning the finiteness of A 0 and B 0 is an immediate consequence of theorem 1 and says as follows Theorem A 0 and B 0 are equal, finite sets and consist of exactly one element x = Γ Γ, i.e. A 0 = B 0 = {Γ Γ}. Proof: Observing Theorem l and the fact that the composition of relations on the sets of all relations on Γ is an associative binary operation, the statement of Theorem has been proved iff the two inclusions (7) and (8) by and Γ Γ (Γ Γ) T ( 0 ) (7) Γ Γ (Γ Γ) T ( 0 ) 1 (8) are true. It is obvious that (7) is true iff the statement [(, X) Γ Γ (X, H) T ( 0 )] (9) (,H) Γ Γ X Γ holds. Choosing X = H for any (, H) Γ Γ we recognize immediately that (7) is true. The proof of (8) is quite the same as the one of (7) such that we are allowed to omit details. This proves Theorem. After this beautiful proof we are going to turn to A and B and want to investigate the question whether these two sets are also finite. 4 The infinity of A and B Figure 5 shows the five graphs, 1,, and H that can be characterized by means of and 1 represented in Figure 6. It shows that the ordered pair (, H) belongs to A (4) = T ( ) T ( ) 1 T ( ) T ( ) 1 7

1 H Figure 5: but it is neither an element of A () nor of A (). A (4) A (i), i =,. This fact implies that Therefore, it is obvious that A has at least the three elements A (), A () and A (4). This fact implies that the situation of A and B is quite different from the situation of A 0 and B 0 such that we suppose that are A and B are even infinite sets. In order to give a proof of this assumption we generalize the example of figure 5 and prove Theorem A and B are infinite sets. Proof: (1) In order to show that A is infinite we suppose to argue by con- 8

1 1 H Figure 6: tradiction that A is finite. Then there necessarily exists a natural number n N, n, such that A = {A (), A (4),..., A (n 1)} with the property that there is an element k {,,..., n 1} for which the equality A (k) = A (n) holds. This is a contradiction to statement (10) by A (n) A (i). (10) i {,,...,n 1} In order to show that (10) is true we argue as follows. (10) is obviously shown, if and only if, we give an ordered pair (, H) A (n) not belonging to A (i) for any i {,,..., n 1}. It turns out that it is even possible to show that there is an ordered pair (, H) A (n) not belonging to A (i) for every i {,,..., n 1}. If n = k, k 1, is even then we generalize the graph from Figure 5. Thus we consider the graphs = (k) and H = n depicted in Figure 7. eneralizing the diagram of Figure 6 we recognize that the ordered pair (, H) = ((k), n ) is an element of but A (n) = T ( ) T ( ) 1... T ( ) T ( ) 1 (11) (, H) = ((k), n ) / A (i) 9

=(k) k quadrilaterals 1 k quadrilaterals 1 k quadrilaterals k- quadrilaterals n- n k quadrilaterals n H = n k quadrilaterals Figure 7: 10

for each i =,,..., n 1. Since the proof of these two statements is a matter of routine checking, we are allowed to omit details. In the case of odd n = k + 1, k 1, we can argue by observing Figure 8 and choosing (, H) = ( (k), n). A routine checking shows that (, H) A (n), but (, H) / A (i) for each i =,,..., n 1. After all, (10) is proved. This implies that our assumption is false and A is infinite. () To argue by contradiction we assume that B is finite. Then there exists an integer n such that B = {B (), B (),..., B (n 1)} with the property that there is an element k {,,..., n 1} for which A (n) is equal to A (k). This equality is a contradiction to statement (1) by B (n) B (i). (1) i {,,...,n 1} We prove (1) by showing the existence of an ordered pair (, H) B (n) not belonging to B (i) for every i {,,..., n 1}. It is useful to distinguish the two cases (a) n 0 mod and (b) n 1 mod. In (a) we take = X(k), n = k +, k 1, k = n and H = X k represented in Figure 9. Observing Figure 9 the proof of (, H) B (n) and (, H) / B (i) for each i =,,..., n 1 is a matter of routine checking. In case (b), n is odd such that n = k + 1, k 1. Then we choose for, H the graphs Y (k) and Y k+1 depicted in Figure 10. It is easy to see that (, H) B (n) and (, H) / B (i) for each i =,,..., n 1. Hence, the assumption is not true. That means that B is infinite. 5 The infinity of A 4 and B 4 In this section we turn toward the elementary relation 1. After having seen that A and B are infinite sets it is not difficult to guess that A 4 and B 4 are also infinite sets. The confirmation of this guess is given in Theorem 4 A 4 and B 4 are infinite sets. 11

1 = (k) k quadrilaterals 1 k quadrilaterals k quadrilaterals k- quadrilaterals n- n k quadrilaterals n H = n k quadrilaterals Figure 8: 1

=X(k) k= n- quadrilaterals X 1 k quadrilaterals X 1 X k quadrilaterals X X k quadrilaterals X k X k k quadrilaterals X k X k+1 k quadrilaterals X k+1 H = X k+ k quadrilaterals Figure 9: 1

Y k= n = Y(k) quadrilaterals Y 1 k quadrilaterals Y 1 Y k quadrilaterals Y k quadrilaterals Y k- Y k k quadrilaterals Y k Y k k quadrilaterals Y k H = Y k+1 k quadrilaterals Figure 10: 14

Proof: (1) It is evident that the main idea of this proof is similar to the idea of the proof of (1) in Theorem. Therefore, we omit details and restrict ourselves to the proof of statement (1) by i {,,...,n 1} A 4 (n) A 4 (i), (1) where n is assumed to be a positive integer with the property that A 4 consists of the elements A 4 (), A 4 (),..., A 4 (n 1). In order to find an ordered pair (, H) A 4 (n) not belonging to each A 4 (i), i =,,..., n 1, we distinguish the two cases n 0 mod and n 1 mod. In the first case n 0 mod, Figure 11 shows the graphs = (n) and H = n satisfying the properties mentioned above. The proof is not quite easy. In spite of his fact we cannot deal with details. In the second case n 1 mod, Figure 1 depicts the corresponding graphs = H(n) and H = H n. It is clear by means of Figure 1 that (, H) A 4 (n) and (, H) / A 4 (i) for each i =,,..., n 1 are true. This completes the proof of (1). Now we conclude by giving a sketch of the proof of (), where () says that B 4 is infinite. Having assumed that B 4 is finite we have to prove the statement i {,,...,n 1} B 4 (n) B 4 (i). (14) by giving an ordered pair (, H) B 4 (n) not belonging to each B 4 (i), i =,,..., n 1, where n is chosen as explained above. If n 0 mod Figure 1 proves that (14) is true for even n and = X(n) and H = X n. If n 1 mod we choose = Y (n) and H = Y n. Figure 14 shows that (, H) B 4 (n) and (, H) / B 4 (i) for each i =,,..., n 1. This completes the proof in the case (). That means that B 4 is infinite. 15

= (n) n triangles and n quadrilaterals 1 1 4 4 n- n- n 4 n- 4 n H = n Figure 11: 16

= H(n) n triangles and n+1 quadrilaterals H 1 H H 1 H H H n- H n- H n- H n H n H = H n n triangles and n quadrilaterals Figure 1: 17

= X(n) n triangles n quadrilaterals X 1 4 X 1 X X X X n- X n- X n X n- X n H = X n n n triangles quadrilaterals Figure 1: 18

Y n = Y(n) n+1 triangles n quadrilaterals Y 1 Y 1 Y Y Y Y n- Y n- Y n- Y n H = Y n Figure 14: 19

eferences [1].Bodendiek, H.Schumacher, K.Wagner. Über elationen auf raphenmengen. Abhandlungen aus dem mathematischen Seminar der Universität Hamburg 51, 1981. [] D.Sawada. An assessment of a selected set of Piagetian-derived operators for the generation of effective learning hierarachies. JME 197. [] C.Thies. Zur Formalisierung des Piagetschen ruppierungsbegriffs unter besonderer Berücksichtigung graphentheoreticher elationen. Dissertation Kiel, 198. [4] K.Wagner,.Bodendiek. raphentheorie I, BI-Wissenschaftsverlag Mannheim 1989. [5] K.Wagner,.Bodendiek. raphentheorie II, BI-Wissenschaftsverlag Mannheim 1990. [6] K.Wagner,.Bodendiek. raphentheorie III, BI-Wissenschaftsverlag Mannheim 199. 0